Perl Weekly Challenge 116: numbers

One way to let me improve my knowledge about Raku (aka Perl 6) is to implement programs in it. Unluckily, I don’t have any production code to implement in Raku yet (sob!). So, why not try solving the Perl Weekly Challenge tasks?

In the following, the assigned tasks for Challenge 110.

PWC 116 - Task 1

This task has been difficult to me: given a number you have to split it in all possible combinations where numbers become subsequent, that is they differ from the previous one by one. As an example, given 1234 you spli it in 1, 2, 3, 4 and every digit differs from the left one by one.
This is my implementation:

sub MAIN( Int $N where { $N >= 10 } ) {
    my @digits = $N.split( '', :skip-empty );
    my $min-length = 1;

    my @numbers;

    my $i = 0;
    my $done = True;
    while $done && $i < @digits.elems {

        

        # first number ever
        @numbers.push: @digits[ $i ] if ! @numbers;
        my $current-number = @numbers[ * - 1 ];

        # compute available next numbers
        my @next-number    = $current-number + 1, $current-number - 1;

        # see if there is room for any of the next
        # numbers in the remaining array of digits
        $done = False;
        for @next-number {
            my $length = $_.Str.chars;
            if $i + $length < @digits.elems {
                my $current = @digits[ $i + 1 .. $i  + $length ].join.Int;
                if $current == $_ {
                    @numbers.push: $current;
                    $i += $length;
                    $done = True;
                    last;
                }
            }
        }
    }

    # all done
    say @numbers if $done;
}




I split the number into its @digits, and then start iterating so that I place the very first number in the list into @numbers, Then I compute how could be the next number, by adding and removing 1, and I search for the number in the subsequent list of available digits.
For example, given 91011 is start with 9, then I compute the next available that would be 10 or 8 and I see if after the 9 there is one or the other, I find 10, so I compute the next one that will be 11 or 9 and search it again.
However, there is no assurance the first number in the list must be one digit only, so this implementation is not fully working.

PWC 116 - Task 2

The second task was easier: find if a number is made by digits that, if summed, provide a pure square of a number.
My implementation is:

sub MAIN( Int $N where { $N >= 10 } ) {
    
    my $sum = $N.split( '' ).map( { $_ * $_ } ).sum;
    say 1 and exit if $sum.sqrt == $sum.sqrt.Int;
    say 0;
}



I start splitting the number into its digits, then I map every digit to its square, and I sum the final list. Next I check if the sqrt of the sum is the same as its Int value, that means the sum is a pure square.

The article Perl Weekly Challenge 116: numbers has been posted by Luca Ferrari on June 8, 2021